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Hello I have following problem: solve equation $\log{(x-5)^2}+\log{(x+6)^2}=2$

and I rewrited this equation as

$2\log{(x-5)}+2\log{(x+6)}=\log{100} \implies 2(\log{(x-5)(x+6))=\log{100}} \implies \log{x^2+x-30}=\log10 \implies x^2+x-40=0 $

and I solved this equation, but I obtained only two solutions and there should be four, so I wonder if it is necessary to create $\log{(x-5)^2(x+6)^2}=\log{100}$ or is a simplier way.

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I assume the question specifies the base of log is $10$? – Macavity Dec 1 '13 at 15:15
yes of course . – Mark Dec 1 '13 at 15:16
up vote 2 down vote accepted

$$\log{(x-5)^2}+\log{(x+6)^2}=2$$ $$\log{((x-5)(x+6))^2}=2$$ $$((x-5)(x+6))^2=10^2=100$$ $$(x-5)(x+6)=\pm10$$Solve from here

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As the logarithm term contains squares, we can allow $x-5,x+6$ to be negative

In fact, $\log b^2=2\log |b|$ for real $b$

As $\log A+\log B=\log (AB),$

$$\implies \log_{10}\{(x-5)(x+6)\}^2=2$$

$$\implies \{(x-5)(x+6)\}^2=10^2=100$$

$$\implies (x-5)(x+6)=\pm10$$

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